A metal crystallizes in a face centred cubic structure. If the edge length of its unit cell is 'a', the closest approach between two atoms in the metallic crystal will be

We have a metal that crystallizes in a face-centred cubic (fcc) lattice. In an fcc unit cell the atoms are in contact along the face diagonal of the cube.

Let the edge length of the cube be $$a$$ and the atomic radius be $$r$$.

For a square face of the cube, the diagonal is obtained from the Pythagoras theorem: the diagonal of a square of side $$a$$ is

$$\sqrt{a^{2}+a^{2}}=\sqrt{2}\,a.$$

Along this face diagonal we encounter three atoms in the order Corner-Face-centre-Corner. Because the atoms touch each other, the total distance from the centre of the first corner atom to the centre of the opposite corner atom equals four radii. Therefore we write the contact condition

$$4r=\sqrt{2}\,a.$$

Solving for the radius, we divide both sides by $$4$$:

$$r=\frac{\sqrt{2}\,a}{4}=\frac{a}{2\sqrt{2}}.$$

The distance of closest approach between two neighbouring atoms is simply twice the radius:

$$\text{Closest approach}=2r=2\left(\frac{a}{2\sqrt{2}}\right)=\frac{a}{\sqrt{2}}.$$

This value corresponds to Option C.

Hence, the correct answer is Option C.